- #26

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But your theta is wrong, at least in my picture it is wrong.

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- Thread starter flyingpig
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- #26

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But your theta is wrong, at least in my picture it is wrong.

- #27

jhae2.718

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- #28

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yeah, you had [tex]<sin\theta, cos\theta,0>[/tex]

- #29

jhae2.718

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Your vector is in <x,y,z> form, right? Then x = rsin(theta), y = rcos(theta), and z = 0.

- #30

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Wait you may be onto something...*goes and fix tex*

- #31

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Okay! Tex is fixed on first page, made some other minor substition fixed. BUt my x^2 still remains..

- #32

jhae2.718

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Use the approach you did in post #3...

I also TeX-ed up my solution; I'll post it in a little bit...

I also TeX-ed up my solution; I'll post it in a little bit...

Last edited:

- #33

SammyS

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The last I looked, you did have inconsistencies between the way θ is drawn in your figure and the way it's used in some of your equations.Okay! Tex is fixed on first page, made some other minor substitution fixed. But my x^2 still remains..

[tex]d\vec{s} = <0,d\vec{y},0>[/tex] should be [tex]d\vec{s} = <0,dy,0>[/tex]

[tex]\hat{r} = <\cos\theta,\sin\theta,0>[/tex] should be [tex]\hat{r} = <\sin\theta,\cos\theta,0>[/tex]

In general [tex]\left|\vec{A}\times\vec{C}\right|=\left|\vec{A}\right|\left|\vec{C}\right|\sin\theta[/tex] so your cross product should have sin θ, not cos θ.

Jumping down to one of your equations for

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-cos\theta d\vec{y} \hat{k}}{x^2 + y^2}[/tex]

After the above corrections and sin θ = x/r, you should have:

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta dy \hat{k}}{x^2 + y^2}[/tex]

[tex]=-\,\frac{x\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x^2 + y^2)^{3/2}}[/tex]

x comes out of the integral because you're integrating over y, not x.

x comes out of the integral because you're integrating over y, not x.

Last edited:

- #34

jhae2.718

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Here's my solution:

View attachment biot.pdf

View attachment biot.pdf

- #35

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This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

- #36

SammyS

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I didn't do the integration, but my solution agrees with this.Here's my solution:

View attachment 33843

But, I do think there is a problem with both solutions.

It's late, so I'll look at it tomorrow.

Below added in

The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.

I intend to redo my solution using (x

.

Last edited:

- #37

jhae2.718

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This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

I used LaTeX :)

MikTeX 2.9 on Windows.

Source:

Code:

```
\documentclass[10pt,letterpaper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fullpage}
\usepackage{graphicx}
\usepackage{float}
\usepackage{xfrac}
\newcommand{\custpic}[4]{\begin{figure}[H]
\begin{center}
\includegraphics[width=#2\textwidth]{#1}
\vspace{-15pt}
\caption{#3}
\vspace{-20pt}
\label{#4}
\end{center}
\end{figure}}
\begin{document}
\custpic{pf5}{.75}{Problem diagram}{pf}
Start with Eq. (\ref{bs}), the Biot-Savart Law:
\begin{equation}\label{bs}
d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{s} \times \hat{r}}{r^2}
\end{equation}
From the diagram:
\begin{align*}
r = \sqrt{x^2 + y^2} \\
x = r\sin(\theta) \\
y = r\cos(\theta) \\
d\vec{s} = d\vec{y}
\end{align*}
Substitute into Eq. (\ref{bs}):
\begin{equation}
d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{y} \times \hat{r}}{x^2 + y^2}
\end{equation}
Taking the cross product, $id\vec{y} \times \hat{r} = dy\sin(\theta) (-\hat{k})$. Substitute in:
\begin{equation}
d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{dy\sin(\theta)}{x^2 + y^2} (-\hat{k})
\end{equation}
From above, $\sin(\theta) = \dfrac{x}{r}$:
\begin{equation}
d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation}
Find $\vec{B}$ by taking the integral over the infinite line:
\begin{equation}
\vec{B} = \frac{\mu_0 i}{4\pi}\int\limits_{-\infty}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation}
\begin{equation}
\vec{B} = 2 \frac{\mu_0 i}{4\pi}\int\limits_{0}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation}
Integrate (I used Wolfram$|$Alpha) to get:
\begin{equation}
\vec{B} = 2 \frac{\mu_0 i}{4\pi}\left[\frac{y}{x\sqrt{x^2 + y^2}}\right]_{0}^{\infty} (-\hat{k})
\end{equation}
The indefinite integral evaluates to:
\begin{equation}
\vec{B} = \frac{\mu_0 i}{2\pi x} (-\hat{k})
\end{equation}
We can confirm this using Ampere's Law (left as an exercise to the reader...)
\end{document}
```

- #38

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- #39

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- #40

SammyS

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Simply use x...

The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.

...

[tex]\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}[/tex]

I think you may find it handy to evaluate the integral from y

- #41

jhae2.718

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You need to run latex on it.

Simply use x_{0}for x, and use (y_{0}- y) for y.

[tex]\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}[/tex]

I think you may find it handy to evaluate the integral from y_{0}to y .

That is an excellent point, and is something I just glossed over.

- #42

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WHat dose that mean "run latex" over it...?

- #43

jhae2.718

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Code:

`pdflatex filename.tex`

- #44

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I am guessing LaTeX is not the same thing as MikTeX lol

- #45

jhae2.718

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Try a LaTeX editor like TeXniCenter, Texmaker, etc to actually write/compile the code if you don't like the CLI.

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